y = (sin 2x + cot x + 2)2 at x = π/2

Given:


y = (sin2x + cotx + 2)2at x =


First, we have to find of given function, f(x),i.e, to find the derivative of f(x)


(xn) = n.xn – 1


The Slope of the tangent is


y = (sin2x + cotx + 2)2


= 2(sin2x + cotx + 2)2 – 1(sin2x) + (cotx) + (2)}


= 2(sin2x + cotx + 2)(cos2x) + ( – cosec2x) + (0)}


(sinx) = cosx


(cotx) = – cosec2x


= 2(sin2x + cotx + 2)(2cos2x – cosec2x)


Since, x =


= 2 (sin2() + cot() + 2)(2cos2() – cosec2())


= 2 (sin() + cot() + 2) (2cos() – cosec2())


= 2 (0 + 0 + 2)(2( – 1) – 1)


sin() = 0, cos() = – 1


cot() = 0,cosec() = 1


= 2(2)( – 2 – 1)


= 4 – 3


= – 12


The Slope of the tangent at x = is – 12


The Slope of the normal =


The Slope of the normal =


The Slope of the normal =


The Slope of the normal =


1